Problem:
A 4-bit serial-in parallel-out shift register is initially set to 1111. The data 1010 is applied to the input. After 3 clock cycles, what will be the output?

Solution:
To solve this problem, we need to understand how a serial-in parallel-out shift register works and how the data is shifted through the register.

Serial-in Parallel-out Shift Register:
A serial-in parallel-out shift register is a type of shift register where data is shifted in serially (one bit at a time) and then parallelly (all bits at once) outputted. It consists of flip-flops connected in series, with each flip-flop storing one bit of data.

Initial State:
The given 4-bit serial-in parallel-out shift register is initially set to 1111. This means that all the flip-flops in the register are initially storing the value 1.

Data Input:
The data 1010 is applied to the input of the shift register. This means that the first clock cycle will shift in the bit 1, the second clock cycle will shift in the bit 0, the third clock cycle will shift in the bit 1, and the fourth clock cycle will shift in the bit 0.

Clock Cycles:
After 3 clock cycles, the output of the shift register will be determined by the bits that have been shifted in. Let's analyze the state of the register after each clock cycle:

1. After the first clock cycle: The data input 1 is shifted in, and the register becomes 11110.
2. After the second clock cycle: The data input 0 is shifted in, and the register becomes 11100.
3. After the third clock cycle: The data input 1 is shifted in, and the register becomes 11010.

Output:
After 3 clock cycles, the output of the shift register will be the parallel output of the register, which is 1101. Therefore, the correct answer is option 'B' - 0101.